Fifty-six percent of Internet users have seen an online group come together to help a person or community solve a problem whereas only 25% have left an online group because of unpleasant interaction. Develop a 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem (to 4 decimals).

95% confidence interval for the proportion of internet users who say online groups have helped solved a problem is between a lower limit of (0.65 - 0.9349/√n) and an upper limit of (0.65 + 0.9349/√n).

Step-by-step explanation:

Confidence interval for a proportion is given as p +/- margin of error (E)

p is the proportion of internet users who say online groups have helped solved a problem = 65% = 0.65

Let the number of internet users be represented by n

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to infinity degrees of freedom and 5% significance level is 1.96

E = critical value × sqrt[p(1-p) ÷ n] = 1.96 × sqrt[0.65(1-0.65) ÷ n] = 1.96 × sqrt(0.2275 ÷ n) = 1.96×0.477/√n = 0.9349/√n

Lower limit of proportion = p - E = 0.65 - 0.9349/√n

Upper limit of proportion = p + E = 0.65 + 0.9349/√n

95% confidence interval is (0.65-0.9349/√n, 0.65+0.9349/√n)

answer: haha (: i’ll go see what it is

step-by-step explanation: